How can I sort a List alphabetically?

Question

I have a List<String> object that contains country names. How can I sort this list alphabetically?

Answer 1

Assuming that those are Strings, use the convenient static method sort:

Collections.sort(listOfCountryNames)

Answer 2

Solution with Collections.sort

If you are forced to use that List, or if your program has a structure like

• Create List
• Add some country names
• sort them once
• never change that list again

then Thilos answer will be the best way to do it. If you combine it with the advice from Tom Hawtin - tackline, you get:

java.util.Collections.sort(listOfCountryNames, Collator.getInstance());

Solution with a TreeSet

If you are free to decide, and if your application might get more complex, then you might change your code to use a TreeSet instead. This kind of collection sorts your entries just when they are inserted. No need to call sort().

Collection<String> countryNames = 
    new TreeSet<String>(Collator.getInstance());
countryNames.add("UK");
countryNames.add("Germany");
countryNames.add("Australia");
// Tada... sorted.

Side note on why I prefer the TreeSet

This has some subtle, but important advantages:

• It's simply shorter. Only one line shorter, though.
• Never worry about is this list really sorted right now becaude a TreeSet is always sorted, no matter what you do.
• You cannot have duplicate entries. Depending on your situation this may be a pro or a con. If you need duplicates, stick to your List.
• An experienced programmer looks at TreeSet<String> countyNames and instantly knows: this is a sorted collection of Strings without duplicates, and I can be sure that this is true at every moment. So much information in a short declaration.
• Real performance win in some cases. If you use a List, and insert values very often, and the list may be read between those insertions, then you have to sort the list after every insertion. The set does the same, but does it much faster.

Using the right collection for the right task is a key to write short and bug free code. It's not as demonstrative in this case, because you just save one line. But I've stopped counting how often I see someone using a List when they want to ensure there are no duplictes, and then build that functionality themselves. Or even worse, using two Lists when you really need a Map.

Don't get me wrong: Using Collections.sort is not an error or a flaw. But there are many cases when the TreeSet is much cleaner.

Answer 3

You can create a new sorted copy using Java 8 Stream or Guava:

// Java 8 version
List<String> sortedNames = names.stream().sorted().collect(Collectors.toList());
// Guava version
List<String> sortedNames = Ordering.natural().sortedCopy(names); 

Another option is to sort in-place via Collections API:

Collections.sort(names);

Answer 4

Better late than never! Here is how we can do it(for learning purpose only)-

import java.util.List;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;

class SoftDrink {
    String name;
    String color;
    int volume; 

    SoftDrink (String name, String color, int volume) {
        this.name = name;
        this.color = color;
        this.volume = volume;
    }
}

public class ListItemComparision {
    public static void main (String...arg) {
        List<SoftDrink> softDrinkList = new ArrayList<SoftDrink>() ;
        softDrinkList .add(new SoftDrink("Faygo", "ColorOne", 4));
        softDrinkList .add(new SoftDrink("Fanta",  "ColorTwo", 3));
        softDrinkList .add(new SoftDrink("Frooti", "ColorThree", 2));       
        softDrinkList .add(new SoftDrink("Freshie", "ColorFour", 1));

        Collections.sort(softDrinkList, new Comparator() {
            @Override
            public int compare(Object softDrinkOne, Object softDrinkTwo) {
                //use instanceof to verify the references are indeed of the type in question
                return ((SoftDrink)softDrinkOne).name
                        .compareTo(((SoftDrink)softDrinkTwo).name);
            }
        }); 
        for (SoftDrink sd : softDrinkList) {
            System.out.println(sd.name + " - " + sd.color + " - " + sd.volume);
        }
        Collections.sort(softDrinkList, new Comparator() {
            @Override
            public int compare(Object softDrinkOne, Object softDrinkTwo) {
                //comparision for primitive int uses compareTo of the wrapper Integer
                return(new Integer(((SoftDrink)softDrinkOne).volume))
                        .compareTo(((SoftDrink)softDrinkTwo).volume);
            }
        });

        for (SoftDrink sd : softDrinkList) {
            System.out.println(sd.volume + " - " + sd.color + " - " + sd.name);
        }   
    }
}

Answer 5

In one line, using Java 8:

list.sort(Comparator.naturalOrder());

Answer 6

Unless you are sorting strings in an accent-free English only, you probably want to use a Collator. It will correctly sort diacritical marks, can ignore case and other language-specific stuff:

Collections.sort(countries, Collator.getInstance(new Locale(languageCode)));

You can set the collator strength, see the javadoc.

Here is an example for Slovak where Š should go after S, but in UTF Š is somewhere after Z:

List<String> countries = Arrays.asList("Slovensko", "Švédsko", "Turecko");

Collections.sort(countries);
System.out.println(countries); // outputs [Slovensko, Turecko, Švédsko]

Collections.sort(countries, Collator.getInstance(new Locale("sk")));
System.out.println(countries); // outputs [Slovensko, Švédsko, Turecko]

Answer 7

Use the two argument for of Collections.sort. You will want a suitable Comparator that treats case appropriate (i.e. does lexical, not UTF16 ordering), such as that obtainable through java.text.Collator.getInstance.

Answer 8

Here is what you are looking for

listOfCountryNames.sort(String::compareToIgnoreCase)

Answer 9

more simply you can use method reference.

 list.sort(String::compareTo);

Answer 10

By using Collections.sort(), we can sort a list.

public class EmployeeList {

    public static void main(String[] args) {
        // TODO Auto-generated method stub

        List<String> empNames= new ArrayList<String>();

        empNames.add("sudheer");
        empNames.add("kumar");
        empNames.add("surendra");
        empNames.add("kb");

        if(!empNames.isEmpty()){

            for(String emp:empNames){

                System.out.println(emp);
            }

            Collections.sort(empNames);

            System.out.println(empNames);
        }
    }
}

output:

sudheer
kumar
surendra
kb
[kb, kumar, sudheer, surendra]

Answer 11

You can use the following line

Collections.sort(listOfCountryNames, String.CASE_INSENSITIVE_ORDER)

It is similar to the suggestion of Thilo, but will not make a difference between upper and lowercase characters.

Answer 12

descending alphabet:

List<String> list;
...
Collections.sort(list);
Collections.reverse(list);

Answer 13

Java 8 ,

countries.sort((country1, country2) -> country1.compareTo(country2));

If String's compareTo is not suitable for your need, you can provide any other comparator.

Answer 14

Same in JAVA 8 :-

//Assecnding order
listOfCountryNames.stream().sorted().forEach((x) -> System.out.println(x));

//Decending order
listOfCountryNames.stream().sorted((o1, o2) -> o2.compareTo(o1)).forEach((x) -> System.out.println(x));

Answer 15

public static void sortByAlphabetCountry(List<Employee> listCountry) {
    listCountry.sort((o1, o2) -> {
        return o1.getName().compareTo(o2.getName());
    });
}

Answer 16

//Here is sorted List alphabetically with syncronized
package com.mnas.technology.automation.utility;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.Iterator;
import java.util.List;

import org.apache.log4j.Logger;
/**
* 
* @author manoj.kumar
*/
public class SynchronizedArrayList {
static Logger log = Logger.getLogger(SynchronizedArrayList.class.getName());
@SuppressWarnings("unchecked")
public static void main(String[] args) {

List<Employee> synchronizedList = Collections.synchronizedList(new ArrayList<Employee>());
synchronizedList.add(new Employee("Aditya"));
synchronizedList.add(new Employee("Siddharth"));
synchronizedList.add(new Employee("Manoj"));
Collections.sort(synchronizedList, new Comparator() {
public int compare(Object synchronizedListOne, Object synchronizedListTwo) {
//use instanceof to verify the references are indeed of the type in question
return ((Employee)synchronizedListOne).name
.compareTo(((Employee)synchronizedListTwo).name);
}
}); 
/*for( Employee sd : synchronizedList) {
log.info("Sorted Synchronized Array List..."+sd.name);
}*/

// when iterating over a synchronized list, we need to synchronize access to the synchronized list
synchronized (synchronizedList) {
Iterator<Employee> iterator = synchronizedList.iterator();
while (iterator.hasNext()) {
log.info("Sorted Synchronized Array List Items: " + iterator.next().name);
}
}

}
}
class Employee {
String name;
Employee (String name) {
this.name = name;

}
}